exact

[logistic_guy]

Solve.

\(\displaystyle 2xy \ dx + (x^2 - 1) \ dy = 0\)

 

[khansaheb]

Solve.

\(\displaystyle 2xy \ dx + (x^2 - 1) \ dy = 0\)

\(\displaystyle \frac{dy}{y} = \frac{2 * x}{1-x^2} {dx}\)​

​

your turn........continue.....​

 

[logistic_guy]

Let

\(\displaystyle M(x,y) = 2xy\)
\(\displaystyle N(x,y) = x^2 - 1\)

Since \(\displaystyle \frac{\partial M}{\partial y} = 2x = \frac{\partial N}{\partial x}\)

There is a function \(\displaystyle f(x,y) = C\) such that

\(\displaystyle \frac{\partial f}{\partial x} = 2xy\)


\(\displaystyle \frac{\partial f}{\partial y} = x^2 - 1\)


If we integrate the first equation with respect to \(\displaystyle x\), we get:

\(\displaystyle \int \frac{\partial f}{\partial x} \ \partial x = \int 2xy \ dx\)


\(\displaystyle f(x,y) = x^2y + g(y)\)

where \(\displaystyle g(y)\) is the constant of integration.

If we take the derivative of \(\displaystyle f(x,y)\) with respect to \(\displaystyle y\), we get:

\(\displaystyle \frac{\partial f}{\partial y} = x^2 + g'(y)\)


\(\displaystyle x^2 - 1 = x^2 + g'(y)\)

This gives:

\(\displaystyle g'(y) = -1\)

If we integrate this with respect to \(\displaystyle y\), we get:

\(\displaystyle \int dg = \int -1 \ dy\)

\(\displaystyle g(y) = -y + c_1\)

Then,

\(\displaystyle f(x,y) = x^2y - y + c_1 = C\)

The solution to the differential equation is:

\(\displaystyle f(x,y) = C\)

Or

\(\displaystyle x^2y - y + c_1 = C\)

Or

\(\displaystyle x^2y - y = C - c_1\)

Let \(\displaystyle c = C - c_1\), then

\(\displaystyle x^2y - y = c\)

Or

\(\displaystyle y(x) = \textcolor{blue}{\frac{c}{x^2 - 1}}\)